Question: The following equation is true for all real values of $n$ for which the expression on the left is defined, and $B$ is a polynomial expression. $\dfrac{n^2+12n+27}{3n^3+9n^2} \div \dfrac{2n^2+18n}{B}=1$ What is $B$ ? $B=$
Explanation: The left side of the equation is a quotient of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting quotient on the left side should cancel out completely. In order to solve for $B$, let's divide the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $n^2+12n+27$, of the dividend can be factored as $(n+9)(n+3)$ by using the sum-product pattern. The denominator, $3n^3+9n^2$, of the dividend can be factored as $3n^2(n+3)$ by factoring out $3n^2$. The numerator, $2n^2+18n$, of the divisor can be factored as $2n(n+9)$ by factoring out $2n$. Now the quotient looks as follows: $\dfrac{(n+9)(n+3)}{3n^2(n+3)} \div \dfrac{2n(n+9)}{B}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{(n+9)(n+3)}{3n^2(n+3)} \div \dfrac{2n(n+9)}{B} \\\\\\ &= \dfrac{(n+9)(n+3)}{3n^2(n+3)} \cdot \dfrac{B}{2n(n+9)} &\text{Flip the divisor} \\\\\\ &= \dfrac{(n+9)(n+3) \cdot B}{3n^2(n+3) \cdot 2n(n+9)} &\text{Multiply across.}\\\\\\ &= \dfrac{{\cancel{(n+9)}}{\cancel{(n+3)}} \cdot B}{3n^2{\cancel{(n+3)}} \cdot 2n{\cancel{(n+9)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{B}{6n^3} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{B}{6n^3}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $B=6n^3$.